466
Suspension Bridge.
polygon does not properly close, the arrows may not be in the
right direction, and a new examination must be made.
Next tak,e the point <2, and draw the triangle F A, A E, E F in
the manner shewn at j. EF should measure 15 cwts. Finally,
'draw the polygon EA, AC, CD, DE for the point ^, as shewn at
K, making CD = 13*9 cwts. and D E = 10 cwts. And the stresses
in the members may now be measured off, marking -h for com-
pression, and - for tension:
-f in AF = 14*5 cwts.
4- in AE = 104 cwts,
- in AC = 9*5 cwts.
Thick and thin lines also represent compression and tension
respectively.
Suspension Bridge Chain.A free uniform rope or chain
hangs in a catenary curve, which is, however, so nearly like a
parabola that the latter is always substituted for simplicity in
practice. Taking the chain in Fig. 424, supposed weightless, but
with loads at even distances as shewn, the forces at L and B are
necessary to keep equilibrium, and the chains will be in tension as
shewn by the arrows. Supposing reactions to be 3^ each, the
triangles ABC, A CD, AEF, &c., are drawn in succession. Then
the distribution of load may be found for CD, DE, EF, &c., and
the stresses in the chain also measured.
Warren Girder with Symmetrical Loads.Mrst,
Distributed on lower boom (Fig. 425). The cells are equilateral
triangles, and the girder has been much used for American
bridges.' Loads being i, i, i, reactions are i^-fij, and the
force i\ at j causes compressions in H B and H A, but tensions in
AJ and AB. The force diagrams are drawn for points i, 2, 3, 4,
5, &c., and the total diagram is given in the figures JKGDA.
Measuring the latter, we find the stresses to be as follows :
In A H and H G = 173 4-
B A and G F =? 173
CB and FE'= 0-58 4-
DC and ED = 0*58 -
In HB and HF = 173 +
HD = 2-31 +
AJ and GK = o'86 -
CM and LE = 2*02